Maximize XOR subsequence possible by equidistant elements from both ends

Given an array A[] of size N, find the Maximum Xor Subsequence such that both A [ i ] and A [ N – i – 1 ] belong to this subsequence, where i ranges between [0, N – 1].

Examples:

Input: N = 8, A [ ] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 13
Explanation:
Maximum Xor Subsequence is {1, 3, 4, 5, 6, 8}

Input: N = 5, A [ ] = {3, 2, 6, 5, 4}
Output: 6
Explanation:
Maximum Xor Subsequence is {3, 2, 6, 5, 4}

Approach:
Since, both A[i] and A[N-i-1] should be present in the same subsequence, pair them up and then find maximum xor sum. For each valid index i, calculate (A[i] ^ A[N-i-1]) and store it in new array X. Size of this new array will be N/2.

Naive Solution:
The simplest approach after creating array X[] for this problem is to recursively generate all subsequences of X and find the maximum xor subsequence. 

Following is the recursive definition of maxXorSubseq(X, N, i):

maxXorSubseq ( X, N, i ) = MAX ( X[ i ] ^ maxXorSubseq ( X, N, i + 1 ), maxXorSubseq (X, N, i + 1) )

The total possible combinations will be 2^N.
Time Complexity: O(2^N)

Below is the implementation of the above approach:

C++

// C++ implementation
// of the above approach
#include <bits/stdc++.h>
using namespace std;

// Returns maximum xor
int maxXorSubseq(vector<int> &x, int n,
                                 int i)
{
    if(i == n)
       return 0;

    return max(x[i] ^ maxXorSubseq(x, n,
                                   i + 1),
                      maxXorSubseq(x, n,
                                   i + 1));
}

vector<int> compute(vector<int> a, int n)
{
    vector<int> x;

    // Calculate a[i]^a[n-i-1]
    for(int i = 0; i < n / 2; i++)
       x.push_back(a[i] ^ a[n - i - 1]);

    // If n is odd
    if(n & 1)
       x.push_back(a[n / 2]);

    return x;
}

// Driver code
int main()
{
    int n = 8;
    vector<int> a = { 1, 2, 3, 4,
                      5, 6, 7, 8 };

    // Getting new array x
    vector<int> x = compute(a, n);

    int mxXor = maxXorSubseq(x, x.size(), 0);

    cout << (mxXor);
    return 0;
}

// This code is contributed by mohit kumar 29

Java

// Java implementation
// of the above approach
import java.util.*;

class GFG{

// Returns maximum xor
static int maxXorSubseq(List<Integer> x, int n,
                                         int i)
{
    if(i == n)
    return 0;

    return Math.max(x.get(i) ^ maxXorSubseq(x, n,
                                            i + 1),
                               maxXorSubseq(x, n,
                                            i + 1));
}

static List<Integer> compute(List<Integer> a, int n)
{
    List<Integer> x = new ArrayList<Integer>();

    // Calculate a[i]^a[n-i-1]
    for(int i = 0; i < n / 2; i++)
        x.add(a.get(i) ^ a.get(n - i - 1));

    // If n is odd
    if((n & 1) == 1)
        x.add(a.get(n / 2));

    return x;
}

// Driver code
public static void main(String[] args)
{
    int n = 8;
    List<Integer> a = Arrays.asList( 1, 2, 3, 4,
                                     5, 6, 7, 8 );

    // Getting new array x
    List<Integer> x = compute(a, n);

    int mxXor = maxXorSubseq(x, x.size(), 0);

    System.out.println((mxXor));
}
}

// This code is contributed by offbeat

Python3

# Python3 implementation
# of the above approach

# Returns maximum xor
def maxXorSubseq(x, n, i):

    if(i == n):
        return 0

    return max(
        x[i]^maxXorSubseq(
        x, n, i + 1), maxXorSubseq(
        x, n, i + 1))

def compute(a, n):

    x = []

    # Calculate a[i]^a[n-i-1]
    for i in range(n//2):
        x.append(a[i]^a[n-i-1])

    # If n is odd
    if(n&1):
        x.append(a[n//2])

    return x

# Driver code
if __name__ =="__main__":

    n = 8
    a = [1, 2, 3, 4, 5, 6, 7, 8]

    # Getting new array x
    x = compute(a, n)

    mxXor = maxXorSubseq(x, len(x), 0)

    print(mxXor)

C#

// C# implementation
// of the above approach
using System;
using System.Collections.Generic;
class GFG {

    // Returns maximum xor
    static int maxXorSubseq(List<int> x, int n, int i)
    {
        if(i == n)
        return 0;

        return Math.Max(x[i] ^ maxXorSubseq(x, n, i + 1), maxXorSubseq(x, n, i + 1));
    }

    static List<int> compute(List<int> a, int n)
    {
        List<int> x = new List<int>();

        // Calculate a[i]^a[n-i-1]
        for(int i = 0; i < n / 2; i++)
            x.Add(a[i] ^ a[n - i - 1]);

        // If n is odd
        if((n & 1) == 1)
            x.Add(a[n / 2]);

        return x;
    }

  static void Main() {
        int n = 8;
        List<int> a = new List<int>{ 1, 2, 3, 4, 5, 6, 7, 8 };

        // Getting new array x
        List<int> x = compute(a, n);

        int mxXor = maxXorSubseq(x, x.Count, 0);

        Console.WriteLine((mxXor));
  }
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

    // Javascript implementation
    // of the above approach

    // Returns maximum xor
    function maxXorSubseq(x, n, i)
    {
        if(i == n)
        return 0;

        return Math.max(x[i] ^ maxXorSubseq(x, n, i + 1),
        maxXorSubseq(x, n, i + 1));
    }

    function compute(a, n)
    {
        let x = [];

        // Calculate a[i]^a[n-i-1]
        for(let i = 0; i < parseInt(n / 2, 10); i++)
            x.push(a[i] ^ a[n - i - 1]);

        // If n is odd
        if((n & 1) == 1)
            x.push(a[parseInt(n / 2, 10)]);

        return x;
    }

    let n = 8;
    let a = [ 1, 2, 3, 4, 5, 6, 7, 8 ];

    // Getting new array x
    let x = compute(a, n);

    let mxXor = maxXorSubseq(x, x.length, 0);

    document.write((mxXor));

</script>

Output:

13

Time Complexity: O(2ⁿ)

Auxiliary Space: O(N)

Efficient Approach:
Considering the above implementation, the following is a partial recursion tree for input X = [9, 5, 1]

In the above partial recursion tree, maxXorSubseq({1}) is being solved four times. On drawing the complete recursion tree, it can be observed that there are many subproblems that are solved again and again. So this problem has Overlapping Subproblems and recomputation of the same subproblems can be avoided by using Memoization. 

For memoization, create an array dp [ ] array and store:

dp[i] = max(x[i] ^ maxXorSubseq(x, n, i + 1), maxXorSubseq(x, n, idx + 1)

This avoids recalculation for previously calculated indices thus optimizing computational complexity.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
vector<int> dp;

// Returns maximum xor sum
int maxXorSubseq(vector<int> x, int n, int idx)
{
    if (idx == n)
    {
        return 0;
    }

    // If already precomputed
    if (dp[idx] != -1)
    {
        return dp[idx];
    }
    int ans = 0;
    ans = max(ans, x[idx] ^ maxXorSubseq(x, n, idx + 1));
    ans = max(ans, maxXorSubseq(x, n, idx + 1));

    // Store the maximum
    dp[idx] = ans;
    return ans;
}

vector<int> compute(int a[],int n)
{
    vector<int> x;

    // Calculate a[i]^a[n-i-1]
    for(int i = 0; i < n / 2; i++)
    {
        x.push_back(a[i] ^ a[n - i - 1]);
    }

    //  If n is odd
    if ((n & 1) != 0)
    {
        x.push_back(a[n / 2]);
    }
    return x;
}

// Driver code
int main()
{
    int n = 8;
    int a[] = { 1, 2, 3, 4, 5, 6, 7, 8 };

    // Getting new array x
    vector<int> x = compute(a, n);

    // Initialize dp array
    for(int i = 0; i < x.size(); i++)
    {
        dp.push_back(-1);
    }
    int mxXor = maxXorSubseq(x, x.size(), 0);
    cout << mxXor << endl;
    return 0;
}

// This code is contributed by divyesh072019.

Java

// Java implementation of the above approach
import java.io.*;
import java.util.*;

class GFG{

static Vector<Integer> dp = new Vector<Integer>();

// Returns maximum xor sum
static int maxXorSubseq(Vector<Integer> x,
                        int n, int idx)
{
    if (idx == n)
    {
        return 0;
    }

    // If already precomputed
    if (dp.get(idx) != -1)
    {
        return dp.get(idx);
    }
    int ans = 0;
    ans = Math.max(ans, x.get(idx) ^
                   maxXorSubseq(x, n, idx + 1));
    ans = Math.max(ans,
                   maxXorSubseq(x, n, idx + 1));

    // Store the maximum
    dp.set(idx,ans);
    return ans;
}

static Vector<Integer> compute(int[] a,int n)
{
    Vector<Integer> x = new Vector<Integer>();

    // Calculate a[i]^a[n-i-1]
    for(int i = 0; i < n / 2; i++)
    {
        x.add(a[i] ^ a[n - i - 1]);
    }

    //  If n is odd
    if ((n & 1) != 0)
    {
        x.add(a[n / 2]);
    }
    return x;
}

// Driver code
public static void main(String[] args)
{
    int n = 8;
    int[] a = { 1, 2, 3, 4, 5, 6, 7, 8 };

    // Getting new array x
    Vector<Integer> x = compute(a, n);

    // Initialize dp array
    for(int i = 0; i < x.size(); i++)
    {
        dp.add(-1);
    }
    int mxXor = maxXorSubseq(x, x.size(), 0);

    System.out.println(mxXor);
}
}

//  This code is contributed by avanitrachhadiya2155

Python3

# Python3 implementation
# of the above approach

# Returns maximum xor sum
def maxXorSubseq(x, n, idx):

    if(idx == n):
        return 0

    # If already precomputed
    if(dp[idx]!=-1):
        return dp[idx]

    ans = 0

    ans = max(
        ans, x[idx] ^ maxXorSubseq(
                    x, n, idx + 1))

    ans = max(ans, maxXorSubseq(
                    x, n, idx + 1))

    # Store the maximum
    dp[idx] = ans

    return ans

def compute(a, n):

    x = []

    # Calculate a[i]^a[n-i-1]
    for i in range(n//2):
        x.append(a[i]^a[n-i-1])

    # If n is odd
    if(n&1):
        x.append(a[n//2])

    return x

# Declared dp[] array globally
dp = []

# Driver code
if __name__ =="__main__":

    n = 8
    a =[1, 2, 3, 4, 5, 6, 7, 8]

    # Getting new array x
    x = compute(a, n)

    # Initialize dp array
    dp = [-1 for i in range(len(x))]

    mxXor = maxXorSubseq(x, len(x), 0)

    print(mxXor)

C#

// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG
{
    static List<int> dp = new List<int>();

    // Returns maximum xor sum
    static int maxXorSubseq(List<int> x, int n, int idx)
    {
        if (idx == n)
        {
            return 0;
        }

        // If already precomputed
        if(dp[idx] != -1)
        {
            return dp[idx];
        }
        int ans = 0;
        ans = Math.Max(ans, x[idx]^maxXorSubseq(x, n, idx + 1));
        ans = Math.Max(ans, maxXorSubseq(x, n, idx + 1));

        // Store the maximum
        dp[idx] = ans;
        return ans;
    }
    static List<int> compute(int[] a,int n)
    {
        List<int> x = new List<int>();

        // Calculate a[i]^a[n-i-1]
        for(int i = 0; i < n / 2; i++)
        {
            x.Add(a[i] ^ a[n - i - 1]);
        }

        //  If n is odd
        if((n & 1) != 0)
        {
            x.Add(a[n / 2]);
        }
        return x;
    }

    // Driver code
    static public void Main ()
    {
        int n = 8;
        int[] a = { 1, 2, 3, 4, 5, 6, 7, 8 };

        // Getting new array x
        List<int> x = compute(a, n);

        // Initialize dp array
        for(int i = 0; i < x.Count; i++)
        {
            dp.Add(-1);
        }
        int mxXor = maxXorSubseq(x, x.Count, 0);
        Console.WriteLine(mxXor);
    }
}

// This code is contributed by rag2127

Javascript

<script>
    // Javascript implementation of the above approach

    let dp = [];

    // Returns maximum xor sum
    function maxXorSubseq(x, n, idx)
    {
        if (idx == n)
        {
            return 0;
        }

        // If already precomputed
        if(dp[idx] != -1)
        {
            return dp[idx];
        }
        let ans = 0;
        ans = Math.max(ans, x[idx]^maxXorSubseq(x, n, idx + 1));
        ans = Math.max(ans, maxXorSubseq(x, n, idx + 1));

        // Store the maximum
        dp[idx] = ans;
        return ans;
    }

    function compute(a, n)
    {
        let x = [];

        // Calculate a[i]^a[n-i-1]
        for(let i = 0; i < parseInt(n / 2, 10); i++)
        {
            x.push(a[i] ^ a[n - i - 1]);
        }

        //  If n is odd
        if((n & 1) != 0)
        {
            x.push(a[n / 2]);
        }
        return x;
    }

    let n = 8;
    let a = [ 1, 2, 3, 4, 5, 6, 7, 8 ];

    // Getting new array x
    let x = compute(a, n);

    // Initialize dp array
    for(let i = 0; i < x.length; i++)
    {
      dp.push(-1);
    }
    let mxXor = maxXorSubseq(x, x.length, 0);
    document.write(mxXor);


// This code is contributed by suresh07.
</script>

Output:

13

Time complexity: O(N)

Auxiliary Space: O(N)